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Unformatted text preview: Lecture 8 Andrei Antonenko February 19, 2003 1 Matrix equations and the inverse 1.1 Discussion of the algorithm  Part 2 Last time we proved the following result: Lemma 1.1. 1. If the square matrix A is invertible, then its RREF is the identity matrix. 2. If we can reduce the matrix A by elementary row operations to the identity matrix, i.e. if its RREF is identity matrix, then this algorithm gives us A 1 B in the right half of the augmented matrix. Actually, the opposite assertion of this lemma is also true. Lemma 1.2. Let RREF of a square matrix A be the identity matrix. Then A is invertible. Proof. Lets consider the process of reducing A to its RREF. It can be done by elementary row operations with matrices E 1 , E 2 , ... , E s , i.e. ( E s E s 1 E 1 ) A = I . From this equality we see that the product of matrices E s E s 1 E 1 satisfy the definition of the inverse for A . So, from these 2 lemmas we get the interesting result, which is the main result about invertible matrices so far: Theorem 1.3. The matrix A is invertible if and only if its RREF is the identity matrix. 2 Vector spaces In this lecture we will introduce a new algebraic structure which is one of the most important structure in linear algebra. This would be a set with 2 operations addition of its elements and multiplication of numbers by its elements. 1 Definition 2.1. Let k be any field. We didnt study fields so far, so those who are not familiar with them can just treat the letter k as another notation for R . A set V is called vector space if there defined an operation of addition of elements of V such that v,w V v + w V , and an operation of multiplication of elements of k by elements of V (often called scalar multiplication ) such that k k v V kv V , and the following axioms are satisfied:...
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 Spring '03
 Andant
 Linear Algebra, Vector Space, Ring

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